Question: The differentiable functions $x$ and $y$ are related by the following equation: $y^2=x^2-5$ Also, $\dfrac{dy}{dt}=2.1$. Find $\dfrac{dx}{dt}$ when $y=-2$ and $x>0$.
Solution: Let's start by differentiating the equation $y^2=x^2-5$ with respect to $t$. $\begin{aligned} y^2&=x^2-5 \\\\ 2y\cdot\dfrac{dy}{dt}&=2x\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that ${\dfrac{dy}{dt}=2.1}$, and we want to find $\dfrac{dx}{dt}$ when ${y=-2}$. We also need to find the corresponding value of ${x}$. To do that, we can plug ${y=-2}$ into the relating equation: $\begin{aligned} {y}^2&={x}^2-5 \\\\ ({-2})^2&={x}^2-5 \\\\ 9&=x^2 \\\\ {x}&{=\pm3} \end{aligned}$ Since we know that $x>0$, that means ${x=3}$. Let's plug ${y=-2}$, ${x=3}$, and ${\dfrac{dy}{dt}=2.1}$ into the equation we obtained: $\begin{aligned} 2{y}\cdot{\dfrac{dy}{dt}}&=2{x}\cdot\dfrac{dx}{dt} \\\\ 2({-2})\cdot{2.1}&=2({3})\cdot\dfrac{dx}{dt} \\\\ -8.4&=6\cdot \dfrac{dx}{dt} \\\\ -1.4&=\dfrac{dx}{dt} \end{aligned}$ In conclusion, when $y=-2$ and $x>0$, the value of $\dfrac{dx}{dt}$ is $-1.4$.